Many elements show very strong similarities to each other.<span>For example, lithium (Li), sodium (Na), and potassium (K) are all soft, very reactive metals.
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Answer:
Here's what I get
Explanation:
I followed the instructions and got the diagram below.
Answer:
See the answer below.
Explanation:
Fire has three major components:
- Heat
- Smoke
- Gases ( in form of CO, CO2 etc)
If the victim had died as a result of the fire, he/he would have inhaled smoke and hot gases from the fire. These components would have resulted in traces of burns and soot deposition in the trachea and lungs as well as traces of CO in the blood of the victim.
If the analysis of the victim's corpse does not reflect some of the results above, it can be effectively concluded that the victim has been dead before the fire.
<em>The single most important indicator of death by the fire would be the presence of CO in the blood of the victim's corpse. All others might be to a less significant degrees.</em>
Answer : The amount of heat released, 45.89 KJ
Solution :
Process involved in the calculation of heat released :

Now we have to calculate the amount of heat released.
where,
Q = amount of heat released = ?
m = mass of water = 27 g
= specific heat of liquid water = 4.184 J/gk
= specific heat of solid water = 2.093 J/gk
= enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole
conversion :
Now put all the given values in the above expression, we get
(1 KJ = 1000 J)
Therefore, the amount of heat released, 45.89 KJ
Answer:
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Explanation:

Moles of mercury(II) acetate = 
Moles of sodium dichromate = 
According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :
of mercury(II) acetate
This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.
According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :
of mercury(II) dichromate
Mass of 0.045906 moles of mercury(II) dichromate:
0.045906 mol × 316.59 g/mol = 14.533 g
14.533 grams of solid precipitate of mercury(II) dichromate will form.